A 40 cm wire having a mass of  3.2 g  is stretched between two fixed supports 40.05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz.   If the area of cross-section of the wire is 1.0 mm2 , then young’s modulus of the wire is ____×1011 N/m2

Given  : L=40 c.m , m=3.2 g , ΔL=40.05−40=0.05 c.m , f1=220 Hz , A=1 mm2   Young's Modulud : Y = TA∆LL ⇒T=AYΔLL=AY0.0540=AY800 ⇒T=AY800=10−6×Y800 Linear Mass Density : μ=m/L=3.2×10−340×10−2=8×10−3 kg/m  Fundamental Frequency : f1=12LTμ ⇒220=12×0.410−6×Y800×8×10−3 ⇒176=Y6.4×106 ⇒Y=6.4×1762×106=1.98×1011 N/m2